Differential equation problem

  • Thread starter yoamocuy
  • Start date
  • #1
41
0
differential equation problem :(

Homework Statement


I'm given a function acceleration a=-1.5*s, where s is a position. I need to find a in terms of t.


Homework Equations





The Attempt at a Solution


I know that a is also equal to d^2*s/dt^2. Therefore d^2*s/dt^2 is equal to -1.5*s. By dividing by s and multiplying by dt^2 I get d^2*s/s=-1.5*dt^2. At this point I'm not sure what to do. If I can figure out how to get rid of the dt^2 and the d^2*s, then I can probably solve the rest of the problem. I imagine I need to integrate but am not sure how that would work out.
 

Answers and Replies

  • #2
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,621
7


Homework Statement


I'm given a function acceleration a=-1.5*s, where s is a position. I need to find a in terms of t.


Homework Equations





The Attempt at a Solution


I know that a is also equal to d^2*s/dt^2. Therefore d^2*s/dt^2 is equal to -1.5*s. By dividing by s and multiplying by dt^2 I get d^2*s/s=-1.5*dt^2. At this point I'm not sure what to do. If I can figure out how to get rid of the dt^2 and the d^2*s, then I can probably solve the rest of the problem. I imagine I need to integrate but am not sure how that would work out.
You cannot simply 'divide' and 'multiply' differentials like that. I know many non-mathematical sciences courses will tell you that it is fine to do so, but it is simply wrong.

I see that you are attempting to separate the variables, however, this method cannot generally be used for second order differential equations. What methods have you learnt for solving second order, homogeneous ODE's?
 
  • #3
41
0


Well this is actually just a physics class so we haven't learned any methods for solving differential equations. I've taken calculus classes but won't be taking differential equations until next semester so I've just been looking through notes online trying to figure out how to solve this question.
 
  • #4
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,621
7


Well this is actually just a physics class so we haven't learned any methods for solving differential equations. I've taken calculus classes but won't be taking differential equations until next semester so I've just been looking through notes online trying to figure out how to solve this question.
The standard method of solving such problems is to take an Ansatz of the form

[tex]s = Ae^{\lambda t}[/tex]

and substitute that into the ODE.

I'm sorry that I don't know of any good online references for DE's, but I'm sure someone here will be able to suggest a suitable reference.
 
  • #5
41
0


Ok, thanks for the help. I'll keep searching and working with it.
 
  • #6
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,621
7


Ok, thanks for the help. I'll keep searching and working with it.
If you would like to post your working, I'd be more than happy to help you with it. All you need to do is substitute that Astatz I gave you into the ODE and you should find that you have a rather simple equation to solve.
 
  • #7
41
0


Ok, I got an equation that looks like s^2*e^((lambda)*t)=-1.5, but when I try to solve for the roots it isn't possible. Am I still completely off on my attempt?
 
  • #8
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,621
7


Ok, I got an equation that looks like s^2*e^((lambda)*t)=-1.5, but when I try to solve for the roots it isn't possible. Am I still completely off on my attempt?
I'm not sure how you've managed to get that. Can you work out what

[tex]\frac{d^2 s}{dt^2} = \frac{d^2}{dt^2} Ae^{\lambda t}[/tex]

is?
 
  • #9
41
0


Ug I dont even know what I'm supposed to be solving for there. On the sites I've been looking at, it looks like they are just taking a Laplace transform and multiplying each differntial by e^-((lambda)*t)
 
  • #10
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,621
7


Ug I dont even know what I'm supposed to be solving for there. On the sites I've been looking at, it looks like they are just taking a Laplace transform and multiplying each differntial by e^-((lambda)*t)
I was simply asking what is the second derivative of [itex]Ae^{\lambda t}[/itex] with respect to t?

Of course, you can use Laplace transforms if you like, but it is much more straightforward (if a little inelegant) to us an Anstatz.
 
  • #12
41
0


I took the second derivitive of that and got A*(lambda)^2*e^((lambda)*t). Would lambda be -2 for that?

I also tried using Laplace transforms and ended up with s^2+1.5=0, which has imaginary roots.
 
  • #13
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,621
7


I took the second derivitive of that and got A*(lambda)^2*e^((lambda)*t). Would lambda be -2 for that?
You're on the right lines. So let's take a look at the over all equation,

[tex]A\lambda^2 e^{\lambda t} = -1.5Ae^{\lambda t}[/tex]

Hence,

[tex]\lambda^2 = -1.5[/tex]

Do you agree?
 
  • #14
41
0


Yea, I agree with that, but if I solve for lambda I get j1.22. What can I do with that?
 
  • #15
Hootenanny
Staff Emeritus
Science Advisor
Gold Member
9,621
7


Yea, I agree with that, but if I solve for lambda I get j1.22. What can I do with that?
Simple, you have determined lambda, so you now have the [general] solution,

[tex]s\left(t\right) = A\exp\left(i\sqrt{1.5}t\right)[/tex]

That is it. You have now found the general solution to the ODE. You can use Euler's relation to write it in a 'nicer' form, but you have found a valid solution.
 

Related Threads on Differential equation problem

  • Last Post
Replies
4
Views
486
  • Last Post
Replies
3
Views
856
  • Last Post
Replies
3
Views
583
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
5
Views
592
  • Last Post
Replies
3
Views
831
Top